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I just cited myself. (mander.xyz)

submitted 1 year ago by fossilesque@mander.xyz to c/science_memes@mander.xyz

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[-] Valthorn@feddit.nu 97 points 1 year ago* (last edited 1 year ago)

x=.9999...

10x=9.9999...

Subtract x from both sides

9x=9

x=1

There it is, folks.

[-] barsoap@lemm.ee 72 points 1 year ago* (last edited 1 year ago)

Somehow I have the feeling that this is not going to convince people who think that 0.9999... /= 1, but only make them madder.

Personally I like to point to the difference, or rather non-difference, between 0.333... and ⅓, then ask them what multiplying each by 3 is.

[-] BugleFingers@lemmy.world 1 points 9 months ago

The thing is 0.333... And 1/3 represent the same thing. Base 10 struggles to represent the thirds in decimal form. You get other decimal issues like this in other base formats too

(I think, if I remember correctly. Lol)

[-] SkyezOpen@lemmy.world 1 points 1 year ago

Oh shit, don't think I saw that before. That makes it intuitive as hell.

[-] DeanFogg@lemm.ee 1 points 1 year ago

Cut a banana into thirds and you lose material from cutting it hence .9999

[-] wholookshere 3 points 1 year ago

That’s not how fractions and math work though.

[-] Blum0108@lemmy.world 49 points 1 year ago

I was taught that if 0.9999... didn't equal 1 there would have to be a number that exists between the two. Since there isn't, then 0.9999...=1

[-] wieson@feddit.org 3 points 1 year ago

Not even a number between, but there is no distance between the two. There is no value X for 1-x = 0.9~

We can't notate 0.0~ ....01 in any way.

[-] Shampiss@sh.itjust.works 25 points 1 year ago

Divide 1 by 3: 1÷3=0.3333...

Multiply the result by 3 reverting the operation: 0.3333... x 3 = 0.9999.... or just 1

0.9999... = 1

[-] ArchAengelus@lemmy.dbzer0.com 6 points 1 year ago

In this context, yes, because of the cancellation on the fractions when you recover.

1/3 x 3 = 1

I would say without the context, there is an infinitesimal difference. The approximation solution above essentially ignores the problem which is more of a functional flaw in base 10 than a real number theory issue

[-] Shampiss@sh.itjust.works 25 points 1 year ago* (last edited 1 year ago)

The context doesn't make a difference

In base 10 --> 1/3 is 0.333...

In base 12 --> 1/3 is 0.4

But they're both the same number.

Base 10 simply is not capable of displaying it in a concise format. We could say that this is a notation issue. No notation is perfect. Base 10 has some confusing implications

[-] chaonaut@lemmy.world 10 points 1 year ago

This seems to be conflating 0.333...3 with 0.333... One is infinitesimally close to 1/3, the other is a decimal representation of 1/3. Indeed, if 1-0.999... resulted in anything other than 0, that would necessarily be a number with more significant digits than 0.999... which would mean that the ... failed to be an infinite repetition.

[-] yetAnotherUser@discuss.tchncs.de 4 points 1 year ago* (last edited 1 year ago)

Unfortunately not an ideal proof.

It makes certain assumptions:

That a number 0.999... exists and is well-defined
That multiplication and subtraction for this number work as expected

Similarly, I could prove that the number which consists of infinite 9's to the left of the decimal separator is equal to -1:

...999.0 = x
...990.0 = 10x

Calculate x - 10x:

x - 10x = ...999.0 - ...990.0
-9x = 9
x = -1

And while this is true for 10-adic numbers, it is certainly not true for the real numbers.

[-] Valthorn@feddit.nu 2 points 1 year ago* (last edited 1 year ago)

While I agree that my proof is blunt, yours doesn't prove that .999... is equal to -1. With your assumption, the infinite 9's behave like they're finite, adding the 0 to the end, and you forgot to move the decimal point in the beginning of the number when you multiplied by 10.

x=0.999...999

10x=9.999...990 assuming infinite decimals behave like finite ones.

Now x - 10x = 0.999...999 - 9.999...990

-9x = -9.000...009

x = 1.000...001

Thus, adding or subtracting the infinitesimal makes no difference, meaning it behaves like 0.

Edit: Having written all this I realised that you probably meant the infinitely large number consisting of only 9's, but with infinity you can't really prove anything like this. You can't have one infinite number being 10 times larger than another. It's like assuming division by 0 is well defined.

0a=0b, thus

a=b, meaning of course your ...999 can equal -1.

Edit again: what my proof shows is that even if you assume that .000...001≠0, doing regular algebra makes it behave like 0 anyway. Your proof shows that you can't to regular maths with infinite numbers, which wasn't in question. Infinity exists, the infinitesimal does not.

[-] yetAnotherUser@discuss.tchncs.de 3 points 1 year ago

Yes, but similar flaws exist for your proof.

The algebraic proof that 0.999... = 1 must first prove why you can assign 0.999... to x.

My "proof" abuses algebraic notation like this - you cannot assign infinity to a variable. After that, regular algebraic rules become meaningless.

The proper proof would use the definition that the value of a limit approaching another value is exactly that value. For any epsilon > 0, 0.999.. will be within the epsilon environment of 1 (= the interval 1 ± epsilon), therefore 0.999... is 1.

[-] sp3tr4l@lemmy.zip 2 points 1 year ago

The explanation I've seen is that ... is notation for something that can be otherwise represented as sums of infinite series.

In the case of 0.999..., it can be shown to converge toward 1 with the convergence rule for geometric series.

If |r| < 1, then:

ar + ar² + ar³ + ... = ar / (1 - r)

Thus:

0.999... = 9(1/10) + 9(1/10)² + 9(1/10)³ + ...

= 9(1/10) / (1 - 1/10)

= (9/10) / (9/10)

= 1

Just for fun, let's try 0.424242...

0.424242... = 42(1/100) + 42(1/100)² + 42(1/100)³

= 42(1/100) / (1 - 1/100)

= (42/100) / (99/100)

= 42/99

= 0.424242...

So there you go, nothing gained from that other than seeing that 0.999... is distinct from other known patterns of repeating numbers after the decimal point.